Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $t = \dfrac{-9q + 72}{q^2 - 64} \times \dfrac{8q + 64}{q + 6} $
Answer: First factor the quadratic. $t = \dfrac{-9q + 72}{(q + 8)(q - 8)} \times \dfrac{8q + 64}{q + 6} $ Then factor out any other terms. $t = \dfrac{-9(q - 8)}{(q + 8)(q - 8)} \times \dfrac{8(q + 8)}{q + 6} $ Then multiply the two numerators and multiply the two denominators. $t = \dfrac{ -9(q - 8) \times 8(q + 8) } { (q + 8)(q - 8) \times (q + 6) } $ $t = \dfrac{ -72(q - 8)(q + 8)}{ (q + 8)(q - 8)(q + 6)} $ Notice that $(q - 8)$ and $(q + 8)$ appear in both the numerator and denominator so we can cancel them. $t = \dfrac{ -72(q - 8)\cancel{(q + 8)}}{ \cancel{(q + 8)}(q - 8)(q + 6)} $ We are dividing by $q + 8$ , so $q + 8 \neq 0$ Therefore, $q \neq -8$ $t = \dfrac{ -72\cancel{(q - 8)}\cancel{(q + 8)}}{ \cancel{(q + 8)}\cancel{(q - 8)}(q + 6)} $ We are dividing by $q - 8$ , so $q - 8 \neq 0$ Therefore, $q \neq 8$ $t = \dfrac{-72}{q + 6} ; \space q \neq -8 ; \space q \neq 8 $